If y(theta) = frac{2 cos theta + cos 2 theta} | vedclass

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(D) Given $y(	heta) = \frac{2 \cos 	heta + 2 \cos^2 	heta - 1}{4 \cos^3 	heta - 3 \cos 	heta + 8 \cos^2 	heta - 4 + 5 \cos 	heta + 2}$.Simplify

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given $y(	heta) = \frac{2 \cos 	heta + 2 \cos^2 	heta - 1}{4 \cos^3 	heta - 3 \cos 	heta + 8 \cos^2 	heta - 4 + 5 \cos 	heta + 2}$.Simplifying the denominator: $4 \cos^3 	heta + 8 \cos^2 	heta + 2 \cos 	heta - 2 = 2(2 \cos^3 	heta + 4 \cos^2 	heta + \cos 	heta - 1)$.Factoring the expression,we get $y(	heta) = \frac{2 \cos^2 	heta + 2 \cos 	heta - 1}{(2 \cos 	heta + 2)(2 \cos^2 	heta + 2 \cos 	heta - 1)} = \frac{1}{2(1 + \cos 	heta)}$.At $	heta = \frac{\pi}{2}$,$y = \frac{1}{2(1 + 0)} = \frac{1}{2}$.Now,$y' = \frac{d}{d	heta} [\frac{1}{2}(1 + \cos 	heta)^{-1}] = \frac{1}{2} (-1)(1 + \cos 	heta)^{-2} (-\sin 	heta) = \frac{\sin 	heta}{2(1 + \cos 	heta)^2}$.At $	heta = \frac{\pi}{2}$,$y' = \frac{1}{2(1)^2} = \frac{1}{2}$.Next,$y'' = \frac{d}{d	heta} [\frac{\sin 	heta}{2(1 + \cos 	heta)^2}] = \frac{1}{2} \left[ \frac{\cos 	heta (1 + \cos 	heta)^2 - \sin 	heta (2(1 + \cos 	heta)(-\sin 	heta))}{(1 + \cos 	heta)^4} ight]$.At $	heta = \frac{\pi}{2}$,$y'' = \frac{1}{2} \left[ \frac{0(1)^2 - 1(2(1)(-1))}{1^4} ight] = \frac{1}{2} [2] = 1$.Thus,$y'' + y' + y = 1 + \frac{1}{2} + \frac{1}{2} = 2$.

If $y(\theta) = \frac{2 \cos \theta + \cos 2 \theta}{\cos 3 \theta + 4 \cos 2 \theta + 5 \cos \theta + 2}$,then at $\theta = \frac{\pi}{2}$,$y'' + y' + y$ is equal to:

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